Question

Jake and paul start together to walk 10 kms.Jake walks 1.5 kmph faster than paul and arrives at his destination 1.5hrs earlier than paul.Jakes's speed of walking is

Answer 1

Answer:

2.25 km/h

Explanation:

Let's take,

V1 & V2 be the velocity of walking of Jake and Paul respectively

Here,

V1 = 1.5 km/hr

V2 = ?

S1 & S2 be distance which they will cover

Here,

S1= S2 = 10 km.....(1)

A/c to relation between velocity , distance & time ---

We have,

S1 = V1× T1.....(2)

& S2= V2× T2......(3)

Now,

From equation (1),(2)&(3)

S1= S2

=> V1× T1 = V2× T2.......(4)

Since , a/c to qestion

Jake arrives at his destination 1.5hrs earlier than paul.

Let,

Paul takes time to cover distance( T2)= X hr

Therefore ,

Jake takes time to cover

distance(T1)=1.5X hr

Now ,

From equation (4)

we have,

V1× T1 = V2× T2

=> 1.5×1.5X = V2× X

=> 2.25X = V2×X

[ cancelled out X on both side ]

=> 2.25 = V2

=> V2 = 2.25

Therefore,

The Required answer is 2.25 km/hr

........Thank You........

Answer 2

Answer:

2.25 Km/h

Explanation:

hope you understand